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Chapter 2: Problem 21

Find dy/dx by implicit differentiation. $$ x e^{2 y}-x^{3}+2 y=5 $$

Short Answer

Expert verified
\(\frac{dy}{dx} = \frac{3x^2 - e^{2y}}{2x e^{2y} + 2}\)

Step by step solution

01

Differentiate both sides of the equation with respect to x

First, let's differentiate both sides of the given equation \(x e^{2y} - x^{3} + 2y = 5\) with respect to x. We should note that \(y\) is a function of \(x\), so when differentiating terms containing \(y\), we should apply the chain rule.
02

Apply Chain Rule to terms containing y

The chain rule states that when differentiating a function of a function, \(\frac{d}{dx}\big[f(g(x))\big] = f'(g(x)) \cdot g'(x)\). Now, we need to apply this rule when differentiating any terms containing \(y\).
03

Differentiate left side of the equation

For simplicity, we differentiate each term in the equation separately: \(1. \frac{d}{dx} (xe^{2y})\): Apply chain rule and product rule. \(\frac{d}{dx} (xe^{2y}) = e^{2y} \cdot \frac{d}{dx} (x) + x \cdot \frac{d}{dx}(e^{2y})\) \(= e^{2y} + x \cdot \frac{d}{d(2y)}(e^{2y})\cdot \frac{d(2y)} {d(y)}\cdot \frac{d(y)}{d(x)}\) \(= e^{2y} + 2x e^{2y} \frac{dy}{dx}\) 2. \(\frac{d}{dx} (-x^{3})\) \(\frac{d}{dx} (-x^{3}) = -3x^{2}\) 3. \(\frac{d}{dx}(2y)\): Apply chain rule. \(\frac{d}{dx}(2y) = 2 \cdot \frac{dy}{dx} = 2\frac{dy}{dx}\) Now, we combine the results: \[\frac{d}{dx}\big (xe^{2y} - x^{3} + 2y\big) = e^{2y} + 2x e^{2y} \frac{dy}{dx} -3x^{2} + 2\frac{dy}{dx}\]
04

Differentiate right side of the equation

Now, let's differentiate the right side of the equation \(\frac{d}{dx}(5)\). Since 5 is a constant, its derivative with respect to x is 0.
05

Solve for dy/dx

Now, we set the derivatives of both left and right sides equal to each other: \(e^{2y} + 2x e^{2y} \frac{dy}{dx} - 3x^{2} + 2\frac{dy}{dx} = 0\) Next, isolate the terms containing dy/dx: \(2x e^{2y} \frac{dy}{dx} + 2\frac{dy}{dx} = 3x^{2} - e^{2y}\) Now, factor out dy/dx: \(\frac{dy}{dx} (2x e^{2y} + 2) = 3x^{2} - e^{2y}\) Finally, divide by the factor in front of dy/dx term to obtain the result: \[\frac{dy}{dx} = \frac{3x^2 - e^{2y}}{2x e^{2y} + 2}\]

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Chain Rule
The chain rule in calculus helps us to differentiate composite functions. A composite function is simply a function inside another function. The chain rule states that the derivative of a function like this is the derivative of the outer function evaluated at the inner function, times the derivative of the inner function.
  • If you have a function written as f(g(x)), then its derivative will be f'(g(x)) multiplied by g'(x).
  • This is useful when we want to differentiate terms involving 'y' in implicit differentiation because 'y' is treated as a function of 'x'.
In our original exercise, for the term involving 'y', such as \(e^{2y}\), we apply the chain rule because \(y\) itself depends on \(x\). This ensures that we properly account for how changes in \(x\) impact \(y\), and subsequently each term containing \(y\).
Product Rule
The product rule is crucial when dealing with the derivatives of products of two functions. If \(u(x)\) and \(v(x)\) are two differentiable functions, the derivative of their product is given by:
  • \(\frac{d}{dx}[u(x)v(x)] = u'(x)v(x) + u(x)v'(x)\)
This allows us to efficiently differentiate products without expanding them, which is useful in complex equations.
In the initial exercise, for example, the term \(xe^{2y}\) is a product of two functions - \(x\) is one function, and \(e^{2y}\) is another. To differentiate \(xe^{2y}\), we apply the product rule:
  • First, differentiate \(x\), keeping \(e^{2y}\) constant, yielding \(e^{2y}\).
  • Then keep the \(x\) constant and differentiate \(e^{2y}\), applying the chain rule.
This results in the expression \(e^{2y} + 2xe^{2y} \frac{dy}{dx}\), combining both the product and chain rules.
Derivatives
Derivatives represent the rate of change of a function concerning one of its variables. They are a fundamental tool in calculus for solving problems related to the slope of curves and instantaneous rates of change.
Generally, when differentiating:
  • Functions without 'y' directly, like simple polynomials, can be done straightforwardly by applying x-power rules, e.g., \(-x^3\) becomes \(-3x^2\).
  • Functions involving 'y', such as \(2y\), requires implicitly differentiating treating 'y' as a function of 'x,' thus introducing \(\frac{dy}{dx}\).
Derivatives are used to find solutions like critical points and slopes, which help to understand the behavior of functions. In implicit differentiation problems, derivatives are applied systematically to every term to solve for the derivative \(\frac{dy}{dx}\) explicitly.
Calculus Problem Solving
Calculus problem solving typically entails breaking down complex equations using rules like the chain rule and product rule to find solutions.
For implicit differentiation:
  • We start by differentiating both sides of an equation with respect to \(x\), treating \(y\) as an implicit function of \(x\).
  • Each term of the equation is differentiated individually, applying pertinent rules to manage products and composite functions.
  • Equating the differentiated expressions on both sides allows for isolating \(\frac{dy}{dx}\), often involving rearranging and factoring.
Solving these exercises not only requires knowledge of differentiation rules but also algebraic manipulation to arrange and simplify the derivatives to achieve clear expressions for \(\frac{dy}{dx}\). This process is fundamental in many real-world applications where relationships are implicitly defined and we need to find rates of change or slopes directly. It enhances critical thinking in both academic and practical real-world contexts.

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Most popular questions from this chapter

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